Case 27 · logic · Hard · About 6–10 minutes
A copied museum plan lists six passages. Exactly one was invented; the other five are genuine. A maintenance walk provides the key to finding the addition.
The walker started in the Dome, finished at the Exit, and used every genuine passage exactly once. Rooms could be revisited, but no passage could be used twice.
Rooms and listed passages
The room abbreviations are A: Atrium, B: Bronze Room, C: Courtyard, D: Dome, and E: Exit.
The copied plan lists these undirected passages:
A–B, A–C, B–C, B–D, C–D, D–E.
Guaranteed rules
- Exactly one listed passage does not exist. Every other listed passage exists.
- There are no additional passages, hidden doors, one-way restrictions, or ways outside the building.
- The recorded walk from D to E is accurate and traverses each genuine passage once.
- Passing through a room does not consume a passage unless the walker uses that connecting route.
Which listed passage must be the invented one? Supply one complete D-to-E walk through the five remaining passages to confirm that your proposed deletion works.
You can draw a dot for each room and a line for each listed passage. The decorative gallery illustration is not a floor plan and supplies no extra connections. The question asks for a unique false passage, not a unique order for the maintenance walk.
Hint 1 - reveal when you want a nudge
At a room other than the start or finish, every arrival must be paired with a departure. Count the passages meeting each room.
Hint 2 - reveal when you want a nudge
Only D and E should have odd counts. On the copied plan, B and C also have odd counts. One false line must fix both.
Reveal the solution and explanation
B–C is the invented passage. One valid walk is D–B–A–C–D–E.
The copied plan gives passage counts A=2, B=3, C=3, D=3, E=1. In a walk using each genuine passage once, intermediate rooms have even counts: each arrival pairs with a departure. The two different endpoints, D and E, have odd counts.
B and C therefore need to change from odd to even, while the other rooms retain their parity. Removing a passage changes the count at its two endpoints by one. It must connect B and C, so the removed entry is B–C.
The example walk uses all five surviving passages exactly once. Reversing the loop gives another valid order, D–C–A–B–D–E. Multiple walks do not weaken the unique identification of the false passage.
Solve another: The Cabinet of Fifty Mysteries · Browse all 50 cases
